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(a) Show that sequence ${a_n} = \sqrt[n]{{{3^n} + {5^n}}}$ is monotone decreasing
Proof Let ${a_n} = \sqrt[n]{{{3^n} + {5^n}}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^n} + 1} \right]^{\frac{1}{n}}}$ then
${a_k} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}}$
${a_{k + 1}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$
We Know ${\left( {\frac{3}{5}} \right)^k} \geqslant {\left( {\frac{3}{5}} \right)^{k + 1}}$

${\left( {\frac{3}{5}} \right)^k} + 1 \geqslant {\left( {\frac{3}{5}} \right)^{k + 1}} + 1$

${\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{{k + 1}}}} \geqslant {\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$

${\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}} \geqslant {\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{{k + 1}}}} \geqslant $ ${\left[ {{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$

$5{\left[ {{{\left( {\frac{3}{5}} \right)}^k} + 1} \right]^{\frac{1}{k}}} \geqslant 5{\left [{{{\left( {\frac{3}{5}} \right)}^{k + 1}} + 1} \right]^{\frac{1}{{k + 1}}}}$

${a_k} \geqslant {a_{k + 1}}$
So ${a_n} = \sqrt[n]{{{3^n} + {5^n}}} = 5{\left[ {{{\left( {\frac{3}{5}} \right)}^n} + 1} \right]^{\frac{1}{n}}}$ is monotone decreasing . I not sure it True or False ,I hope someone help me thank.

Doretomtom
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2 Answers2

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Let $a=3/5$. We look at $\log(1+a^x)^{1/x}=\frac{\log(1+a^x)}{x}$ for positive $x$.

Differentiate, using the Quotient Rule. The denominator is safely positive. The numerator is $$\frac{x}{1+a^x}(\log a)a^x-\log(1+a^x).$$ This is negative, since $\log a$ is. Since the logarithm is decreasing, so is the original function.

André Nicolas
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Hint: Taking the log-derivative of $5\left(1+(3/5)^n\right)^{1/n}$ yields $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}n}\left(\log(5)+\frac1n\log\left(1+(3/5)^n\right)\right) &=\frac1n\frac{\color{#C00000}{\log(3/5)}(3/5)^n}{1+(3/5)^n}\color{#C00000}{-\frac1{n^2}}\log\left(1+(3/5)^n\right)\end{align} $$

I've just noticed that André's answer uses the quotient rule while mine uses product rule; otherwise, mine is not much different. To give added value to my answer, I feel I should give an answer that does not use calculus.

Bernoulli's Inequality, the rational version of which is proven here, shows $$ \begin{align} \left[1+\left(\frac35\right)^n\right]^{\Large\frac{n+1}{n}} &\ge1+\frac{n+1}{n}\left(\frac35\right)^n\\ &\ge1+\frac35\left(\frac35\right)^n\\ &=1+\left(\frac35\right)^{n+1}\\ \left[1+\left(\frac35\right)^n\right]^{\Large\frac1n} &\ge\left[1+\left(\frac35\right)^{n+1}\right]^{\Large\frac1{n+1}}\\ \left[5^n+3^n\vphantom{3^{n+1}}\right]^{\Large\frac1n} &\ge\left[5^{n+1}+3^{n+1}\right]^{\Large\frac1{n+1}} \end{align} $$

Community
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robjohn
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